Reading Nuclear Equations

Reading Nuclear Equations Video

Hi, and welcome to this video on nuclear reactions. Today, we’re going to delve into the notation used to represent them, and as we do that, we’ll review their definition, how they usually take place, and the common species involved.

Let’s begin with a quick refresher on nuclear reactions and how they differ from the chemical reactions that you’re probably more familiar with.

In chemical reactions, electrons are redistributed between chemical species, breaking and forming new bonds, while the nuclei remain unchanged. This means that the elemental identity of the reactants doesn’t change in the course of a chemical reaction. So for example, if we start with 4 atoms of hydrogen and 2 atoms of oxygen, we’ll end with 4 atoms of hydrogen and 2 atoms of oxygen.

\(2H_{2} + O_{2}\rightarrow 2H_{2}O\)

The atoms might be bonded differently, in new molecules, or have changed ionic states, but because the nucleons (the protons and neutrons) didn’t participate in the reaction, the elemental identity of each atom goes unchanged.

Remember that all atoms of an element have the same atomic number, Z, which is the number of protons. So if that number changes, the elemental identity of the atom changes as well. However, atoms of the same element can have different mass numbers (A), called isotopes, because the number of neutrons can vary without changing the identity of the atom.

In nuclear reactions, the nucleons are involved in the reaction and thus, new elements and isotopes are formed in the process. So to accurately represent a nuclear reaction, we must include specific information about the nucleons in our notation.

For example, consider this nuclear fusion reaction where two heavy isotopes of hydrogen, deuterium (hydrogen-2), and tritium (hydrogen-3), fuse to form helium and a highly energetic neutron.

\(\frac{2}{1}H+\frac{3}{1}H\rightarrow \frac{4}{2}He+\frac{1}{0}n\)


We have maintained the same standard format used for a chemical equation with the reactants on the left side of the arrow and products on the right side of the arrow. But for each species, along with the atomic symbol, we have now included the mass number and atomic number. Because the mass number is the total nucleons, we get an indirect count of the neutrons as well, which is important because different isotopes can significantly impact the energetics of the reaction.

Including this information also helps us check the mass and charge balance of our nuclear reaction. Remember that even though new elements are created, nuclear reactions still need to follow basic principles, like the law of mass conservation.

This becomes slightly more complicated than in chemical reactions because appreciable mass-energy conversion takes place. Consequently, like in this reaction, the amount of energy released will often be included with the products to show the balanced equation.

\(\frac{2}{1}H+\frac{3}{1}H\rightarrow \frac{4}{2}He+\frac{1}{0}n+17.5\text{ MeV}\)


However, more simply, we must conserve the number of nucleons in the reaction. So, even though nucleons are rearranging, we have to account for all of our neutrons and protons, which also ensures that we conserve charge throughout the reaction.

Let’s consider our initial example. Are the nucleons balanced?

\(\frac{2}{1}H+\frac{3}{1}H\rightarrow \frac{4}{2}He+\frac{1}{0}n\)


By adding the atomic numbers on the reactant side, we get \(1+1=2\) protons and on the product side, we get \(2+0=2\). So we started with 2 protons and ended with 2 protons, which is what we want!

Next, add the mass numbers to confirm that we haven’t lost any neutrons. From the reactant side, \(2+3=5\) and from the product side, \(4+1=5\). So we started with 5 nucleons and ended with 5 nucleons. Perfect!

This is also helpful for predicting products. For example, if we have an unknown product, we can solve for its identity by balancing the reaction. Pause the video and work out the other product, \(_{Z}^{A}\textrm{X}\), in the following radioactive decay of radium:

\(_{88}^{226}\textrm{Ra}\rightarrow _{Z}^{A}\textrm{X}+_{4}^{2}\textrm{He}\)


Got it? Let’s check. To solve for , set up the following simple equations.

The sum of the mass numbers of the reactants is equal to the sum of the mass numbers of the products,

\(\sum \left ( A_{reactants} \right )=\sum \left ( A_{products} \right )\)




\(A = 222\)


We also need to set the sum of the atomic numbers of the reactants equal to the sum of the atomic numbers of the products:


\(88 = Z + 2\) \(Z = 86\)


Knowing Z is 86 reveals radium decayed by emitting a helium-4 nuclei and leaving behind a radon atom.

\(_{88}^{226}\textrm{Ra}\rightarrow _{A}^{Z}\textrm{X}+\frac{4}{2}He\)


In these nuclear decay reactions, you’ll also hear the reactant, in this case radium, called the parent and the larger product called the daughter, which in this case is radon.

You may have noticed that in our notation, some of the information is redundant. For instance, given the atomic symbol and mass number, we can figure out the number of neutrons; consequently, you may see the atomic numbers dropped.

More significantly, you’ll sometimes see the reaction collapsed down into this format:

\(T\left ( P,x \right )R\)


This notation is used for nuclear transmutation reactions, in which scientists induce the elemental conversion of a nucleus. In the experimental setup, there is a target species, T, which is bombarded with a projectile, P. This produces an emitted particle, x, and a residual nucleus, R.

Using the notation of a standard chemical reaction, this is the same as saying

\(T+P\rightarrow x+R\)


So, along with being shorter, this notation provides additional information about the experiment and the energy of the particles. Remember that to execute nuclear reactions, we often have to force positively charged particles together, which requires large amounts of energy, and typically produces lots of energy, so knowing which particles are high-energy is relevant to understanding the reaction.

Let’s consider another example.

Beryllium-9 reacts with helium to produce carbon-12 and a neutron.

\(_{9}^{4}\textrm{Be}+_{2}^{4}\textrm{He}\rightarrow _{6}^{12}\textrm{C}+_{1}^{0}\textrm{n}\)


Using the condensed notation, we write this reaction like this:

\(^{9}\textrm{Be}\left (^{4}\textrm{He}, n \right )^{12}\textrm{C}\)


which now includes the information that beryllium was bombarded with high-energy helium particles to produce carbon and high-energy neutrons.

Apart from the condensed notation for the reaction, some particles are so common to nuclear reactions, particularly as emissions in radioactive decay, that scientists have given them their own name. Consequently, to quickly read nuclear reactions, it’s nice to be comfortable with these terms as well.

In this notation, the superscript and subscript prior to the emission symbol is the mass and charge, respectively.


ParticleAbbreviationChargeMass (amu)
Alpha particle\(_{4}^{2}\textrm{He}_{4}^{2}\textrm{He}^{2+}, \text{ }_{4}^{2}\textrm{a},_{4}^{2}\textrm{a}^{2+}, a\)+24.001506
Beta Particle\(_{-}\frac{0}{1}\beta ,\beta ,\beta –\)-10.000549
Positron\(_{+}\frac{0}{1}\beta ,\beta+\)+10.000549
Neutron\(n, \frac{1}{0}\)01.0008665
Photon, gamma rayY, \(\text{ }_{0}^{0}\textrm{Y}\)

Alpha particles are high-energy helium-4 nuclei, two protons, and two neutrons. They have been stripped of their electrons, so they carry a +2 charge. They can be written as \(_{4}^{2}\textrm{He}^{2+}\) , but are often abbreviated to simply α.

Thus, returning to our previous examples,

\(^{9}Be(^{4}He, n)^{12}C\text{ and}_{88}^{226}\textrm{Ra}\rightarrow _{88}^{226}\textrm{Rn}+_{2}^{4}\textrm{He}\)


we now know that \(^{4}_{2}H\) is an alpha particle. So the first reaction would be written like this:

\(_{88}^{226}\textrm{Ra}\rightarrow _{86}^{222}\textrm{Rn}+a\)


and is commonly referred to as the alpha decay of radium.

The second reaction would be written like this:

\(^{9}Be\left ( a, n \right )^{12}C\)


and read as, beryllium was bombarded with alpha particles to produce carbon and high-energy neutrons.

Beta particles are high-energy electrons emitted from the nucleus when a neutron is converted into a proton and denoted as \(_{-1}^{0}\beta or \text{ }\beta ^{-}\).

They are physically equivalent to electrons, e-, but are produced in a specific way, so it’s best not to interchange the notation.

\(_{0}^{1}\textrm{n}\rightarrow _{1}^{1}\textrm{p}+_{-1}^{0}\beta\)


Positrons are positively charged electrons or the antiparticle to electrons and are denoted as \(_{+1}^{0}\beta or\text{ }\beta ^{+}\).

They are formed when a proton in the nucleus is converted to a neutron, which then expels a positron.

\(_{1}^{1}\textrm{p}\rightarrow _{1}^{0}\textrm{n}+_{+1}^{0}\beta\)

And gamma rays are high-energy photons with no mass or charge.

Also remember that n is a neutron and p is a proton.

Alright, we’ve covered the basics of writing nuclear reactions, both the notation and the common abbreviations. You should now have the tools to read and write most nuclear reactions, but we’ve really only scratched the surface of the theory, so don’t worry if you still feel a little overwhelmed. There’s a lot to learn here!


Finally, let’s wrap up with some practice (if you don’t already, get a periodic table handy!). Pause the video and write the equations for the following nuclear reactions. For the second question, write the full equation as well as the shorthand notation for transmutation reactions.

1. Naturally occurring thorium-232 undergoes α decay.

First, write out what you know- Thorium decays into something plus an alpha particle

\(_{90}^{232}\textrm{Th}\rightarrow _{Z}^{A}\textrm{X}+_{2}^{4}\textrm{a}\)

Now, solve for the unknown particle using the following equations



\(\text{ Therefore,}\text{ X}=_{88}^{228}\text{Ra}+_{2}^{4}\alpha\)

\(_{90}^{232}\textrm{Th}\rightarrow _{Z}^{A}\textrm{X}+_{2}^{4}\textrm{a}\)


2. When aluminum foil was bombarded with α particles, phosphorus-30 and neutrons were formed.

Again, write out what you know- aluminum was bombarded with alpha particles to form phosphorous-30 and neutrons.

\(_{13}^{A}\textrm{Al}+_{2}^{4}\textrm{}\alpha \to _{15}^{30}\textrm{P}+_{0}^{1}\textrm{n}\)

Note that we don’t know the mass number of aluminum, but can quickly solve for it.

\(_{13}^{27}\textrm{Al}+_{2}^{4}\textrm{}\alpha\to _{15}^{30}\textrm{P}+_{0}^{1}\textrm{n}\)

And the shorthand notation is \(^{27}Al\)(α,n)\(^{30}P\)


That’s it for this review on nuclear reactions! Thanks for watching, and happy studying!


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by Mometrix Test Preparation | This Page Last Updated: July 21, 2023

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